Mathematical Methods for Physics and Engineering - Matematica.NET

STATISTICSDifferentiating with respect to µ and setting the result equal to zero gives∂ ln L∂µ = 1T V −1 (x − µ1) =0.Thus, the ML estimator is given by∑ˆµ = 1T V −1 x1 T V −1 1 = i,j (V −1 ) ij x j∑i,j (V .−1 ) ijIn the case of uncorrelated errors in measurement, (V −1 ) ij = δ ij /σi 2 and our estimatorreduces to that given in (31.72). ◭In all the examples considered so far, the likelihood function has been effectivelyone-dimensional, either instrinsically or under the assumption that the values ofall but one of the parameters are known in advance. As the following exampleinvolving two parameters shows, the application of the ML method to theestimation of several parameters simultaneously is straightforward.◮In an experiment N measurements x i of some quantity are made. Suppose the random erroron each sample value is drawn independently from a Gaussian distribution of mean zero butunknown standard deviation σ (which is the same for each measurement). Calculate the MLestimates of the true value µ of the quantity being measured and the standard deviation σof the random errors.In this case the log-likelihood function is given byln L(x; µ, σ) =− 1 N∑[ln(2πσ 2 )+ (x ]i − µ) 2.2σ 2 i=1Taking partial derivatives of ln L with respect to µ and σ and setting the results equal tozero at the joint estimate ˆµ, ˆσ, weobtainN∑ x i − ˆµ=0, (31.73)ˆσ 2 i=1N∑ (x i − ˆµ) 2 N∑ 1− =0. (31.74)ˆσ 3 ˆσi=1i=1In principle, one should solve these two equations simultaneously for ˆµ and ˆσ, but in thiscase we notice that the first is solved immediately byˆµ = 1 N∑x i = ¯x,Ni=1where ¯x is the sample mean. Substituting this result into the second equation, we findˆσ = √ 1 N∑(x i − ¯x)N2 = s,i=1where s is the sample standard deviation. As shown in subsection 31.4.3, s is a biasedestimator of σ. The reason why the ML method may produce a biased estimator isdiscussed in the next subsection. ◭1260